Quy đồng mẫu thức các phân thức Bài 14 trang 27 SBT Toán 8 Tập 1

Giải SBT Toán 8 Bài 4: Quy đồng mẫu thức nhiều phân thức

Bài 14 trang 27 SBT Toán 8 Tập 1: Quy đồng mẫu thức các phân thức:

a) $\frac{7x-1}{2x^2+\text{}6x};\frac{5-3x}{x^2-9}$

b) $\frac{x+\text{}1}{x-x^2};\frac{x+\text{}2}{2-4x+\text{}2x^2}$;

c) $\frac{4x^2-3x+\text{}5}{x^3-1};\frac{2x}{x^2+x+\text{}1};\frac{6}{x-1}$;

d) $\frac{7}{5x};\frac{4}{x-2y};\frac{x-y}{8y^2-2x^2}$;

e) $\frac{5x^2}{x^3+\text{}6x^2+\text{}12x+\text{}8};\frac{4x}{x^2+4\text{}x+4};\frac{3}{2x+\text{\hspace{0.17em}}4}$

Lời giải:

a) Ta có: 2x² + 6x = 2x(x + 3); x² – 9 = (x + 3)(x – 3)

Mẫu thức chung: 2x(x + 3)(x – 3)

$\frac{7x-1}{2x^2+\text{}6x}=\frac{7x-1}{2x(x+3)}=\frac{(\text{}7x-1).(x-3)}{2x(x+\text{}3).(x-3)};$

$\frac{5-3x}{x^2-9}=\frac{5-3x}{(x+\text{}3)(x-3)}=\frac{2x(5-3x)}{2x(x+\text{}3)(x-3)}$

b) Ta có: x – x² = x(1 – x); 2 – 4x + 2x² = 2(1 – 2x + x²) = 2(1 – x)²

Mẫu thức chung: 2x(1 – x)²

^{$\begin{array}{l}\frac{x+\text{}1}{x-x^2}=\frac{x+\text{\hspace{0.17em}}1}{x(1-x)}\\ =\frac{(x+\text{}1)\mathrm{.2.}(1-x)}{x(1-x).2(1-x)}=\frac{2(1-x^2)}{2x{(1-x)}^2};\end{array}$}

^{$\frac{x+\text{}2}{2-4x+\text{}2x^2}=\frac{x\text{}+2}{2{(1-x)}^2}=\frac{(x\text{}+2).x}{2x{(1-x)}^2}$}

c) Ta có: x³ – 1 = (x – 1)(x² + x + 1).

Mẫu thức chung: x³ – 1 = (x – 1)(x² + x + 1)

$\frac{4x^2-3x+\text{}5}{x^3-1}=\frac{4x^2-3x+\text{}5}{(x-1).(x^2+\text{\hspace{0.17em}}x+\text{}1)};$

$\frac{2x}{x^2+x+\text{}1}=\frac{2x.(x-1)}{(x^2+x+\text{\hspace{0.17em}}1).(x-1)};$

$\frac{6}{x-1}=\frac{6(x^2+\text{\hspace{0.17em}}x+\text{}1)}{(x-1).(x^2+\text{\hspace{0.17em}}x+\text{\hspace{0.17em}}1)}$

d) Ta có: 8y² – 2x² = 2(4y² – x²) = 2(2y + x)(2y – x)

Mẫu thức chung: 10x(2y + x)(2y – x)

$\begin{array}{l}\frac{7}{5x}=\frac{7.2(2y+x).(2y-x)}{5x.2(2y+x).(2y-x)}\\ =\frac{14(2y+x).(2y-x)}{10x(2y+x).(2y-x)};\end{array}$

$\begin{array}{l}\frac{4}{x-2y}=\frac{-4}{2y-x}=\frac{-40x(2y+\text{}x)}{10x(2y-x)(2y+x)}\\ =\frac{-4.10x(2y+\text{}x)}{(2y-x).10x(2y+x)}\\ =\frac{-40x(2y+\text{}x)}{10x(2y-x)(2y+x)};\end{array}$

$\begin{array}{l}\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2(2y+\text{}x).(2y-x)}\\ =\frac{(x-y).5x}{2(2y+\text{}x).(2y-x).5x}\\ =\frac{5x(x-y)}{10x(2y+\text{}x).(2y-x)};\end{array}$

e) Ta có: x³ + 6x² + 12x + 8 = x³ + 3x².2 + 3.x.2² + 2³ = (x + 2)³

x² + 4x + 4 = (x + 2)²; 2x + 4 = 2(x + 2)

Mẫu thức chung: 2(x + 2)³

^{$\begin{array}{l}\frac{5x^2}{x^3+\text{\hspace{0.17em}}6x^2+\text{\hspace{0.17em}}12x+\text{}8}=\frac{5x^2}{{(x+\text{}2)}^3}\\ =\frac{5x^2.2}{{(x+\text{}2)}^3.2}=\frac{10x^2}{2{(x+\text{}2)}^3};\end{array}$}

^{$\begin{array}{l}\frac{4x}{x^2+\text{\hspace{0.17em}}4x+4}=\frac{4x}{{(x+2)}^2}\\ =\frac{4x\mathrm{.2.}(x+2)}{{(x+2)}^2.2(x+\text{}2)}=\frac{8x(x+2)}{2{(x+2)}^3};\end{array}$}

^{$\begin{array}{l}\frac{3}{2x+\text{\hspace{0.17em}}4}=\frac{3}{2(x+\text{}2)}\\ =\frac{3{(x+\text{}2)}^2}{2(x+\text{}2).{(x+\text{}2)}^2}=\frac{3{(x+\text{}2)}^2}{2{(x+\text{}2)}^3}\end{array}$}